MadSci Network: Physics

Re: Space travel in three dimensions

Area: Physics
Posted By: Kurt Frost, ,N/A
Date: Mon Apr 14 15:34:46 1997
Message ID: 859488683.Ph

Well, I must say, this problem really hurt my head!  :-)  There was a lot 
to think about, as the solution is not as easy as it may seem.......since 
we are talking about going to 98% the speed of light, we (unfortunately) 
have to venture into the crazy, confusing world of Special Relativity!!!  
I have not looked at this topic for some time, nor was I ever much of an 
expert at it, so please bear with me as I do my best to hack my way through
this problem.  To help me get to my answer, I made use of two of my old 
physics texts and one of my old physics profs (Dr. Taylor) - all of which 
are referenced at the end.  Well, lets get to it......

First of all, your two assumptions are correct:
1.  Each subsequent vector v2 and v3, add to the total velocity
2.  v2 and v3 do NOT reduce any previous acceleration.
However, the total sum of the velocities will NEVER be greater than the 
speed of light (this is a BIG no no in conventional physics).  The one 
problem in your question that makes one have to think a little deeper at 
how this problem must be solved, is that a velocity is not a quantity that
can exist by is always in reference to some else. (eg.  an 
airplane moves at a velocity, "relative" to the ground)  So, I had to consider 
what the reference frames in this problem were and where all of these velocity 
measurements would be taken from.

My first assumption is that you have at least a small knowledge of Special 
Relativity. (I will do my best to explain the important concepts pertaining
to the problem, but I will not derive every equation.)  Now, let's set up the 
problem........assume that the velocities of the space ship are measured from 
a stationary reference frame, S.  The ship will then be moving in the S' 
reference frame, at a velocity, v, relative to the S frame.  You mention in 
your problem that the ship first accelerates to maxV, rotates 90 degrees, then 
accelerates to maxV, then rotates 90 degrees (to the first two directions), and 
then accelerates to maxV.  I will just assume that the ship is starting off going 
in the 'x' direction, then rotating into the 'y', and finally into the 'z' directions.  
Then, with these assumptions, the ship's velocity from the S frame can be represented 
by the following equations:

	ux = (ux' + v) / (1 + (v*ux')/c^2)

	uy = (1/gma)*(uy') / (1 + (v*ux')/c^2)

	uz = (1/gma)*(uz') / (1 + (v*ux')/c^2)

		where:	ux 	- velocity of the ship in the x direction, in the S frame
			  	  (similarly for y and z)
			ux' 	- velocity of the ship in the x direction, in the S' frame
  		  	          (similarly for y' and z')
			v    	- velocity of the S' frame (relative to
				  the S frame)
			gma	- (gma is short form for gamma) time
	     	  		  [gma = 1/ (1 - (v^2)/(c^2))^0.5]
			c	- speed of light (300,000,000 m/s 
						       or 3E8 m/s)

(Just to explain these equations a bit.......ux, uy, and uz come out of the 
'Einstein velocity addition rule', which is derived by using 'Lorentz transformations'.  
These equations, along with gma, are the backbone to special relativity.  If you 
want to know more of how these equations come about, I urge you to look up an advanced 
physics text, like the two I referenced at the end.)


Now, if the space ship took off in the x direction, it's velocity as seen 
from S, will be:

	ux = (ux' + v) / (1 + (v*ux')/c^2)

However, there is no velocity of the ship within the S' frame, so ux' = 0.  
[An example of an object moving within its own moving frame would be a stationary 
person measuring (ux) another person 'walking' (ux') on a moving train (v).]

	ux = (0 + v) / (1 + (v*0)/c^2)
	=>   ux = (v) / (1)	=>	| ux = v |
So, the ship looks to be travelling at the same speed in S that it is travelling 
at in S', the x direction (which is v = maxV = 0.98*c).	


If the ship suddenly turns 90 degrees and heads off in the y direction, 
it's new velocity can be written as:

	uy = (1/gma)*(uy') / (1 + (v*ux')/c^2)

The ship is still not moving in the x direction within the S' frame, so ux' = 0.  
However, uy' does not equal zero (like ux'), because any new velocity not along 
the initial direction of motion of S' (i.e. in the y direction) will be a relative 
velocity within this frame, not just the velocity of the frame (as it is for ux').  
Also, when the ship rotates 90 degrees into the y direction and begins to accelerate 
to maxV, or 0.98c, its velocity in the x direction does not change (from ux = maxV = 0.98c).  

Solving:	uy = (1/gma)*(uy')  or  = (uy')/gma

		=>	| uy = v / gma |	(v = 0.98c)

	(uy' = v because the ship has accelerated to its maximum velocity,
	 v, in the y direction.)


The total combined velocity of the ship moving in the x and y directions, Vxy, 
can be calculated using the Pythagorean theorem:

	Vxy = (ux^2 + uy^2)^0.5	  =>	Vxy = (v^2 + (v^2)/(gma^2))^0.5

	but, gma = 1/ (1 - (v^2)/(c^2))^0.5

	so:	Vxy = (v^2 + (v^2)*(1 - (v^2)/(c^2)))^0.5
	          = (v^2*(2 - (v^2)/(c^2))^0.5

Subbing in for v and c:		| Vxy = 2.99E8 m/s |

(which works out to about 99.9968% c, but most importantly, less than the 
speed of light!; without considering special relativity, the answer would 
have surely exceeded the speed of light)


Similar calculations (to those done for y) can now be done for the z direction:
	uz = (1/gma)*(uz')	=>	or   | uz = (uz')/gma = v / gma |

The only difference between the y and z calculation, that I can see, will 
come out in when you calculate 'gma'. [gma = 1/ (1 - (v^2)/(c^2))^0.5]  
This time the time dialation will be due to the ship's combined x and y 
velocity, or Vxy.  So, sub in Vxy for v in the gma calculation.


Finally, to get a final/total velocity of the ship in all three directions, Vxyz, 
you have redo the Pythagorean calculation, using Vxy and uz:

	Vxyz = (Vxy^2 + uz^2)^0.5

Subbing in and solving:

	| Vxyz = 2.99E8 m/s = 99.9968% c |

This is one fast ship.......but still not faster than the speed of light!!!  
Once again, without considering special relativity, the answer would have 
been much larger than the speed of light (by a factor of about 1.7).


Now to answer your second question, as to whether the apparent loss of 
information about v1 and v2 (or ux and uy), after the ship achieves v3, 
has any significance to the previous calculations.  In short, I would say NO.

The reason I say this, is that it does not matter how an object gets to its 
final position or velocity, you can still measure the final results, which 
amount to the total of all of the previous events.  As an example, consider 
an airplane you see flying overhead.  You don't know of all the turns and 
changes in velocity it has gone through before you saw it, but you can still 
measure its current speed, position, and direction.  As far as you are concerned, 
it has been moving in the way you are viewing it, all along......and this does not 
matter, as it amounts to the same as if you could sum all of plane's previous 
information on speed and direction.  You would get the same answer.

It is the same for the space ship.  If I came along later on to view the ship 
with you, after it had completed its turn from x to y and had just turned towards z, 
I would see and measure the same final velocity of the ship as you would.....even 
though you have previous measurements from ux and uy to work with.


Well, I hope this explanation has helped you and not just confused you even more.  
I also hope that I did not make any glaring mistakes in any of my assumptions or 
calculations....remember I have not done this in a while.  On this note I bid you 
farewell and recommend that if you have any other special relativity questions, you 
had better ask me soon before all of this information I crammed in my head starts to 
seep out again!!	:-)

	Kurt Frost


Benson, Harris.  "University Physics". Toronto: John Wiley & Sons, Inc.,

Griffiths, David J.  "Introduction to Electrodynamics".  New Jersey:
	Prentice Hall, 1989.

Taylor, Dr. David W.  Professor of Physics and Astronomy, McMaster
	University, Hamilton, Ontario, Canada.

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